Difference between revisions of "2007 AMC 8 Problems/Problem 19"
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<math>\mathrm{(A)}\ 2 \qquad \mathrm{(B)}\ 64 \qquad \mathrm{(C)}\ 79 \qquad \mathrm{(D)}\ 96 \qquad \mathrm{(E)}\ 131</math> | <math>\mathrm{(A)}\ 2 \qquad \mathrm{(B)}\ 64 \qquad \mathrm{(C)}\ 79 \qquad \mathrm{(D)}\ 96 \qquad \mathrm{(E)}\ 131</math> | ||
| − | == Solution == | + | == Solution 1 == |
| − | Let the smaller of the two numbers be <math>x</math>. Then, the problem states that <math> | + | Let the smaller of the two numbers be <math>x</math>. Then, the problem states that <math> (x+1)+x<100</math>. <math> (x+1)^2-x^2=x^2+2x+1-x^2=2x+1 </math>. <math> 2x+1 </math> is obviously odd, so only answer choices C and E need to be considered. |
| − | <math> 2x+1=131 </math> | + | <math> 2x+1=131 </math> contradicts the fact that <math> 2x+1<100 </math>, so the answer is <math> \boxed{\mathrm{(C)} 79} </math> |
| + | |||
| + | ==Solution 2== | ||
| + | Since for two consecutive numbers <math>a</math> and <math>b</math>, the difference between their squares is <math>a^2-b^2=(a+b)(a-b)</math>, which equals to <math>a+b</math>, because <math>a</math> and <math>b</math> are consecutive. Because they are consecutive, one number must be even, and the other odd. Since the sum of an even and an odd number is always odd, and the sum of <math>a</math> and <math>b</math> is less than 100, you can eliminate all answers except for <math>\boxed{\mathrm{(C)} 79}</math>. | ||
| + | |||
| + | ==Solution 3 == | ||
| + | From the question we can make <math>a^2-b^2</math> and we can let <math>a</math> be <math>b-1</math> where <math>a+b<100</math>. First we factor <math>a^2-b^2</math> to <math>(a+b)(a-b)</math> and then plug in <math>a=b-1</math> to have: <math>(a+a+1)(a-a+1)=(2a+1)(1)=2a+1=a+b</math>. since <math>a+b</math> will always be an odd number, we can narrow the answers down to (C) and (E). Since <math>131</math> would be too big for our range of <math>a+b<100</math>, we would choose <math>\boxed{\mathrm{(C)} 79}</math>. note that this solution would be better if the question was asking for the values of a and b | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/8FGl2vtO1vs Soo, DRMS, NM | ||
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/BrEqmDq82rw | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2007|num-b=18|num-a=20}} | {{AMC8 box|year=2007|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 11:55, 24 December 2024
Contents
Problem
Pick two consecutive positive integers whose sum is less than 
. Square both
of those integers and then find the difference of the squares. Which of the
following could be the difference?
Solution 1
Let the smaller of the two numbers be 
. Then, the problem states that 
. 
. 
is obviously odd, so only answer choices C and E need to be considered.
contradicts the fact that 
, so the answer is 
Solution 2
Since for two consecutive numbers 
and 
, the difference between their squares is 
, which equals to 
, because and are consecutive. Because they are consecutive, one number must be even, and the other odd. Since the sum of an even and an odd number is always odd, and the sum of and is less than 100, you can eliminate all answers except for .
Solution 3
From the question we can make 
and we can let be 
where 
. First we factor to 
and then plug in 
to have: 
. since will always be an odd number, we can narrow the answers down to (C) and (E). Since 
would be too big for our range of , we would choose . note that this solution would be better if the question was asking for the values of a and b
Video Solution
https://youtu.be/8FGl2vtO1vs Soo, DRMS, NM
Video Solution by WhyMath
~savannahsolver
See Also
| 2007 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
